The sum of 8 different positive integers is at least $1+2+3+\cdots+8=36$, so we must have $n \geq 36$. Now $n=36$ satisfies the desired property, since in this case we must have equality - the eight numbers must be $1, \ldots, 8$. And if $n=37$ the eight numbers must be $1,2, \ldots, 7,9$ : if the highest number is 8 then the sum is $36<n$, while if the highest number is more than 9 the sum is $>1+2+\cdots+7+9=37=n$. So the highest number must be 9 , and then the remaining numbers must be $1,2, \ldots, 7$. Thus $n=37$ also has the desired property. However, no other values of $n$ work: if $n>37$ then $\{1,2,3, \ldots, 7, n-28\}$ and $\{1,2, \ldots, 6,8, n-29\}$ are both sets of 8 distinct positive integers whose sum is $n$. So $n=36,37$ are the only solutions.